How to use the IF ALL statement in Python

Your code can in fact be reduced to checking if nums is sorted, e.g.

def checker(nums):
    return sorted(nums) == nums

This does exactly what you expect, e.g.

>>> checker([1, 1, 2, 2, 3])
True
>>> checker([1, 1, 2, 2, 1])
False

Your function can be reduced to this:

def checker(nums):
    return all(i <= j for i, j in zip(nums, nums[1:]))

Note the following:

• zip loops through its arguments in parallel, i.e. nums[0] & nums[1] are retrieved, then nums[1] & nums[2] etc.
• i <= j performs the actual comparison.
• The generator expression denoted by parentheses () ensures that each value of the condition, i.e. True or False is extracted one at a time. This is called lazy evaluation.
• all simply checks all the values are True. Again, this is lazy. If one of the values extracted lazily from the generator expression is False, it short-circuits and returns False.

Alternatives

To avoid the expense of building a list for the second argument of zip, you can use itertools.islice. This option is particularly useful when your input is an iterator, i.e. it cannot be sliced like a list.

from itertools import islice

def checker(nums):
    return all(i <= j for i, j in zip(nums, islice(nums, 1, None)))

Another iterator-friendly option is to use the itertools pairwise recipe, also available via 3rd party more_itertools.pairwise:

# from more_itertools import pairwise  # 3rd party library alternative
from itertools import tee

def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = tee(iterable)
    next(b, None)
    return zip(a, b)

def checker(nums):
    return all(i <= j for i, j in pairwise(nums))

Another alternative is to use a functional approach instead of a comprehension:

from operator import le

def checker_functional(nums):
    return all(map(le, nums, nums[1:]))